Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
Advertisements
उत्तर
Let I = `int_"0"^pi (x"d"x)/(1 + sin x)` .....(i)
= `int_0^pi (pi - x)/(1 + sin(pi - x)) "d"x` ......`["Using" int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x)"d"x]`
= `int_0^pi (pi - x)/(1 + sinx) "d"x` ......(ii)
Adding (i) and (ii), we get
2I = `int_0^pi (x/(1 + sinx) + (pi - x)/(1 + sinx)) "d"x`
= `int_0^pi ((x + pi - x)/(1 + sinx))"d"x`
= `int_0^pi pi/(1 + sin x) "d"x`
= `pi int_0^pi 1/(1 + sinx) "d"x`
= `pi int_0^pi (1.(1 - sinx))/((1 + sinx)(1 - sinx)) "d"x`
= `pi int_0^pi (1 - sinx)/(1 - sin^2x) "d"x`
= `pi int_0^pi (1 - sinx)/(cos^x) "d"x`
= `pi int_0^pi (1/(cos^2x) - sinx/(cos^2x))"d"x`
= `pi int_0^pi (sec^2x - secx tanx)"d"x`
= `pi[tanx - sec]_0^pi`
= `pi[tan pi - tan 0) - (sec pi - sec 0)]`
2I = `pi[0 - (-1 - 1)`
= `pi`(2)
∴ I = `pi`
Hence, I = `pi`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int x^2/((x^2+2)(2x^2+1))dx`
Evaluate: `∫8/((x+2)(x^2+4))dx`
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Integrate the rational function:
`1/(e^x -1)`[Hint: Put ex = t]
Find `int (2cos x)/((1-sinx)(1+sin^2 x)) dx`
Find :
`∫ sin(x-a)/sin(x+a)dx`
Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`
Integrate the following w.r.t. x : `(12x^2 - 2x - 9)/((4x^2 - 1)(x + 3)`
Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`
Integrate the following w.r.t. x : `(3x - 2)/((x + 1)^2(x + 3)`
Integrate the following w.r.t. x : `(1)/(sin2x + cosx)`
Integrate the following with respect to the respective variable : `(6x + 5)^(3/2)`
Integrate the following w.r.t. x: `(x^2 + 3)/((x^2 - 1)(x^2 - 2)`
Integrate the following with respect to the respective variable : `(cos 7x - cos8x)/(1 + 2 cos 5x)`
Integrate the following with respect to the respective variable : `cot^-1 ((1 + sinx)/cosx)`
Integrate the following w.r.t.x: `(x + 5)/(x^3 + 3x^2 - x - 3)`
Integrate the following w.r.t.x : `sqrt(tanx)/(sinx*cosx)`
Evaluate: `int "3x - 2"/(("x + 1")^2("x + 3"))` dx
Evaluate: `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx
If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)
`int (7 + 4x + 5x^2)/(2x + 3)^(3/2) dx`
`int sqrt((9 + x)/(9 - x)) "d"x`
`int "e"^(sin^(-1_x))[(x + sqrt(1 - x^2))/sqrt(1 - x^2)] "d"x`
`int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
`int x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3)) "d"x`
Evaluate:
`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
`int 1/(sinx(3 + 2cosx)) "d"x`
Choose the correct alternative:
`int (x + 2)/(2x^2 + 6x + 5) "d"x = "p"int (4x + 6)/(2x^2 + 6x + 5) "d"x + 1/2 int 1/(2x^2 + 6x + 5)"d"x`, then p = ?
Evaluate `int x^2"e"^(4x) "d"x`
`int 1/(4x^2 - 20x + 17) "d"x`
If `intsqrt((x - 7)/(x - 9)) dx = Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, then A = ______
The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, the denominator becomes eight times the numerator. Find the fraction.
Evaluate: `int (dx)/(2 + cos x - sin x)`
Find : `int (2x^2 + 3)/(x^2(x^2 + 9))dx; x ≠ 0`.
Evaluate:
`int 2/((1 - x)(1 + x^2))dx`
